Uniform electric fields and field strength

Key ideas

An electric field is a region where a charge feels a force. Between two parallel charged plates the field is uniform — the same strength and direction everywhere — and points from the positive plate to the negative plate.

Two horizontal parallel plates, the top positive and the bottom negative, with evenly spaced vertical field lines pointing downward from the positive plate to the negative plate.

Electric field strength

For a uniform field between plates a distance $d$ apart with a voltage $V$ across them:

$E = \frac{V}{d}$

$E$ — electric field strength (V m⁻¹, or equivalently N C⁻¹)
$V$ — voltage (potential difference) between the plates (V)
$d$ — separation of the plates (m)

Force on a charge in the field

Any charge $q$ placed in the field feels a force:

$F = qE$

$F$ — force on the charge (N)
$q$ — size of the charge (C)
$E$ — field strength (N C⁻¹)

A positive charge is pushed in the direction of the field (toward the negative plate); a negative charge is pushed the opposite way.

:::tip Convert units before substituting. Plate separations are usually given in millimetres and charges in micro- or nanocoulombs. A $5.0$ mm gap is $5.0 \times 10^{-3}$ m, and $2.0$ µC is $2.0 \times 10^{-6}$ C. Forgetting the powers of ten is the number-one error here. :::

Worked ExampleField strength and force between plates

Two parallel plates are $4.0$ mm apart with $200$ V across them. Find the field strength, and the force on a $3.0$ µC charge placed between them.

Step 1 — Convert units

$d = 4.0 \text{ mm} = 4.0 \times 10^{-3} \text{ m}, \quad q = 3.0 \times 10^{-6} \text{ C}$

Step 2 — Field strength

$E = \frac{V}{d} = \frac{200}{4.0 \times 10^{-3}} = 5.0 \times 10^{4} \text{ N C}^{-1}$

Step 3 — Force on the charge

$F = qE = (3.0 \times 10^{-6})(5.0 \times 10^{4}) = 0.15 \text{ N}$

Practice question

Two plates $10$ mm apart have a field of $3.0 \times 10^{4}$ N C⁻¹ between them. Find the voltage across them.

Worked solution: rearrange $E = V/d$ → $V = Ed = (3.0 \times 10^{4})(10 \times 10^{-3}) = 300$ V.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Two parallel plates are $5.0$ mm apart with a potential difference of $150$ V across them.

Calculate the electric field strength between them.

Merit

A charge of $4.0$ µC is placed between two parallel plates $2.0$ mm apart with $100$ V across them.

Calculate the force on the charge.

Excellence

Two identical charged plates are held at a fixed voltage. Explain fully what happens to (a) the field strength and (b) the force on a fixed charge between them if the plate separation is doubled.

Uniform electric fields and field strength

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Key ideas

Two horizontal parallel plates, the top positive and the bottom negative, with evenly spaced vertical field lines pointing downward from the positive plate to the negative plate.

Electric field strength

For a uniform field between plates a distance $d$ apart with a voltage $V$ across them:

$E = \frac{V}{d}$

$E$ — electric field strength (V m⁻¹, or equivalently N C⁻¹)
$V$ — voltage (potential difference) between the plates (V)
$d$ — separation of the plates (m)

Force on a charge in the field

Any charge $q$ placed in the field feels a force:

$F = qE$

$F$ — force on the charge (N)
$q$ — size of the charge (C)
$E$ — field strength (N C⁻¹)

A positive charge is pushed in the direction of the field (toward the negative plate); a negative charge is pushed the opposite way.

Worked ExampleField strength and force between plates

Two parallel plates are $4.0$ mm apart with $200$ V across them. Find the field strength, and the force on a $3.0$ µC charge placed between them.

Step 1 — Convert units

$d = 4.0 \text{ mm} = 4.0 \times 10^{-3} \text{ m}, \quad q = 3.0 \times 10^{-6} \text{ C}$

Step 2 — Field strength

$E = \frac{V}{d} = \frac{200}{4.0 \times 10^{-3}} = 5.0 \times 10^{4} \text{ N C}^{-1}$

Step 3 — Force on the charge

$F = qE = (3.0 \times 10^{-6})(5.0 \times 10^{4}) = 0.15 \text{ N}$

Practice question

Two plates $10$ mm apart have a field of $3.0 \times 10^{4}$ N C⁻¹ between them. Find the voltage across them.

Worked solution: rearrange $E = V/d$ → $V = Ed = (3.0 \times 10^{4})(10 \times 10^{-3}) = 300$ V.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Two parallel plates are $5.0$ mm apart with a potential difference of $150$ V across them.

Calculate the electric field strength between them.

Merit

A charge of $4.0$ µC is placed between two parallel plates $2.0$ mm apart with $100$ V across them.

Calculate the force on the charge.

Excellence

Two identical charged plates are held at a fixed voltage. Explain fully what happens to (a) the field strength and (b) the force on a fixed charge between them if the plate separation is doubled.